Alternate current and complex description

1. Jul 2, 2011

Lindsayyyy

Hi everyone I hope I'm in the right board for this:

When it comes to alternate current, a complex description is pretty common as it's apperently easier to calculate with. But I don't understand alternate current to be honest. I have a problem with the basics, especially the following:

I can describe alternate current with sine, for example:

$$I(t)=I_{0}*sin(\omega t)$$

where I_0 is the max amplitude.

but I can also write an alternate current like this:
$$I(t)=I_{0}*e^{i \omega t}$$
i=imaginary unit
but with Euler's formula I can reqrite this to:

$$I(t)=I_{0}*[cos(\omega t)+i*sin(\omega t)]$$

What does the imaginary part tells me here? Both equations express an alternate current, but they are different (to me at least:) )

I'm pretty sure I have a wrong imagination of this stuff. Can someone explain it to me? I hope you get the point where my problems are.

2. Jul 2, 2011

Gordianus

An alternate current has the first temporal dependence ( sin(omega t)) and is always a real number.
The second one is a complex number which is used as an auxiliary variable that allows solving the problem "easily".
Thus, we shouldn't label both of them as I(t) because the two expressions are different.

3. Jul 2, 2011

tiny-tim

Hi Lindsayyyy!

Current and complex current are two completely different things

Current is variable (it depends on t).

Complex current is constant.

From the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=303"

In a steady sinusoidal (AC) circuit of frequency ω, the (instantaneous) voltage and current $V\text{ and }I$ can always be written:
$V =\ V_x\cos\omega t + V_y\sin\omega t$ and $I =\ I_x\cos\omega t + I_y\sin\omega t$​

Then the complex voltage and complex current between any two points are the constants defined as $\mathbf{V} =\ V_x+jV_y\text{ and }\mathbf{I} =\ I_x+jI_y$.​

Last edited by a moderator: Apr 26, 2017
4. Jul 2, 2011

I like Serena

Hey tiny-tim, you've got me confused saying complex current is a constant.
I thought complex current is not so much constant, but its absolute value is constant.

Complex current is still time dependent, just like real current is.
More precisely real current is the "real" part of the complex current.

To the OP, consider for instance a pendulum.

Its displacement is a cosine of the time.
When the bob is in the position of maximum displacement, all its energy is potential energy.
When the displacement is zero, all the energy of the bob is kinetic.
These two alternate with each other, but at every time the total energy is constant.

So we can describe the motion of the pendulum with a complex number.
The real part would be the actual displacement, related to potential energy.
While the imaginary part would indicate the speed of the bob, related to kinetic energy.

If we multiply the complex value with the appropriate constant, the squared absolute value is the total energy.

Last edited: Jul 2, 2011
5. Jul 2, 2011

Lindsayyyy

Thanks for the help thus far. My problem is I don't understand how exactly I should use the complex version and especially for what I need it. The complex one isn't a physical value because we learned physical values are alway real :/ I'm confused :grumpy:

6. Jul 2, 2011

I like Serena

Yes, physical values are always real.
So if you need the physical value from a complex one, you take the "real" part of the complex number (the part that does not have i in it).
(Do you know what I mean with the "real" part?)

You would use the complex version just like the real version.
Meaning you can multiply, add and subtract them.

As for when you need it. It's basically an arbitrary choice.
Both forms work.
But some calculations are less work with complex values.

7. Jul 3, 2011

Lindsayyyy

Yes, I know what you mean by real part. Did I understand it right you say:
If I have an alternate current given as a complexe number. I just take the real part and ignore the imaginary one?

8. Jul 3, 2011

I like Serena

Hmm, it's a little more complex (no pun intended ).

If you would measure the current with an ammeter, you'll measure the real part, so in this case the imaginary part is "ignored".

If you calculate the effect of a capacitor, its impedance is typically given as a complex number, depending on the frequency of the current ($Z = \frac 1 {i \omega C}$).
So if you calculate the voltage from the current through the capacitor with the complex form of Ohm's law (V = I Z), you need to calculate this with complex numbers.
If you "ignore" the imaginary part of the current in this calculation, you'll get the wrong "real" voltage in your answer.

9. Jul 3, 2011

Lindsayyyy

ok, I think I understand it a bit better now, thank you.

10. Jul 3, 2011

Naty1

11. Jul 3, 2011

Lindsayyyy

one more question:

We can write the impendance as:

$$Z=R+iX$$

i:imaginary unit

Now what's the R exactly? Does a capacitor or indctor exist of an ohm resistance R and reactence X or is Z just the sum of the components in an electric circuit?

12. Jul 3, 2011

f95toli

Capacitors are usually described as "ideal" lossless components, but when losses are taken into account it is usually via the so-called ESR parameter (equivalent series resistance). However, this is a "mode", the R does not correspond to anything specific but is just used to model all the non-ideal behaviour of a real capacitor.

However, if you write down the impedance of an inductor it is easy to see that the imaginary part of the impedance is due to the inductance, and the real part (R) is due to the resistance of the wire (which can be significant in large coils).

13. Jul 3, 2011

Lindsayyyy

ok, thanks everyone for the help and patience. I have to do some exercises on this when I'm through with electrostatics.Be prepared to get annoyed

14. Jul 3, 2011

I like Serena

We'll be waiting with anticipation!